Question: $y'=xy^2$ Is $y=\dfrac{2}{9+x^2}$ a solution to the above equation? Choose 1 answer: Choose 1 answer: (Choice A) A Yes (Choice B) B No
Answer: In order to find whether $y=\dfrac{2}{9+x^2}$ is a solution, we need to substitute it into the equation and see if we get equivalent expressions on each side of the equal sign. In addition to substituting for $y$, we need to find the corresponding $y'$ expression to substitute into the equation: $\begin{aligned} y'&=\dfrac{d}{dx}\left[\dfrac{2}{9+x^2}\right] \\\\ &=-\dfrac{4x}{(9+x^2)^2} \end{aligned}$ Now we substitute ${y=\dfrac{2}{9+x^2}}$ and ${y'=-\dfrac{4x}{(9+x^2)^2}}$ into the equation: $\begin{aligned} {y'}&=x{y}^2 \\\\ {-\dfrac{4x}{(9+x^2)^2}}&\stackrel{?}{=}x\left({\dfrac{2}{9+x^2}}\right)^2 \\\\ -\dfrac{4x}{(9+x^2)^2}&\stackrel{?}{=}x\left(\dfrac{4}{(9+x^2)^2}\right) \\\\ -\dfrac{4x}{(9+x^2)^2}&\stackrel{?}{=}\dfrac{4x}{(9+x^2)^2} \\\\ -4&\neq 4 \end{aligned}$ We did not obtain equivalent expressions on each side. In conclusion, no, $y=\dfrac{2}{9+x^2}$ is not a solution to the differential equation.